Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxi = 200001;
long long int gcd(long long int a, long long int b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int satr(long long int i, long long int n) {
  long long int p = i;
  long long int sum = 0;
  while (p > 0) {
    sum += p % 10;
    p /= 10;
  }
  return (i * i + sum * i - n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  double n, h;
  cin >> n >> h;
  double x1 = h / pow(n, 0.5);
  for (int i = 1; i < n; i++) {
    cout << setprecision(100) << x1 * pow(i, 0.5) << " ";
  }
}